Variance of mean from a model $X_t=\mu+a_t-\theta a_{t-1}$ Let $X_t$ be generated by the following model:
$$X_t=\mu+a_t-\theta a_{t-1}$$
where $a_t\sim N(0,1)$ i.i.d.
Let $\bar{X}=(\sum_{t=1}^nX_t)/n$.
Then $\operatorname{var}((\sum_{t=1}^nX_t)/n)=\frac{1}{n^2}\sum_{t=1}^n\operatorname{var}(X_t)+\frac{2}{n^2}\sum_{t=1}^n\sum_{j=1}^{t-1}\operatorname{cov}(X_t,X_j)$.
I am not sure how the last sentence comes out. Can anyone explain a bit?
 A: Actually, the last line is true in general and thus doesn't depend on the form of $X_t$. Indeed, recalling the properties $\mathrm{Var}[\alpha X] = \alpha^2\mathrm{Var}[X]$ and
$
\mathrm{Var}[X+Y] = \mathrm{Var}[X] + \mathrm{Var}[Y] + \mathrm{Cov}[X,Y] + \mathrm{Cov}[Y,X] =$$ \mathrm{Var}[X] + \mathrm{Var}[Y] + 2\,\mathrm{Cov}[X,Y],
$
since $\mathrm{Cov}[X,Y] = \mathrm{Cov}[Y,X]$, we can extend them to
$$
\begin{array}{rcl}
\mathrm{Var}[\overline{X}] 
   &=& \displaystyle
   \mathrm{Var}\left[\frac{1}{n}\sum_{t=1}^n X_t\right] \\
   &=& \displaystyle
   \frac{1}{n^2}\mathrm{Var}\left[\sum_{t=1}^n X_t\right] \\
   &=& \displaystyle
   \frac{1}{n^2}\sum_{t=1}^n\mathrm{Var}[X_t] + \frac{1}{n^2}\sum_{s,t=1}^n\mathrm{Cov}[X_s,X_t] \\
   &=& \displaystyle
   \frac{1}{n^2}\sum_{t=1}^n\mathrm{Var}[X_t] + \frac{2}{n^2}\sum_{t=1}^n\sum_{s=1}^{t-1}\mathrm{Cov}[X_s,X_t] \\
\end{array}
$$
since $\mathrm{Cov}[X_s,X_t] = \mathrm{Cov}[X_t,X_s]$ appears twice in the double sum of the third line.

Addendum.
Given that linear combinations of normal laws (and constants) are also normal, more precisely
$$
\lambda_0 + \lambda_1\mathcal{N}(\mu_1;\sigma_1^2) + \lambda_2\mathcal{N}(\mu_2;\sigma_2^2) \sim \mathcal{N}(\lambda_0 + \lambda_1\mu_1+\lambda_2\mu_2; \lambda_1^2\sigma_1^2+\lambda_2^2\sigma_2^2),
$$
your concrete case leads to $X_t \sim \mathcal{N}(1;1+\theta^2)$ and $\overline{X} \sim \mathcal{N}\left(1;\frac{1+\theta^2}{n}\right)$, hence $\mathrm{Var}[\overline{X}] = \frac{1+\theta^2}{n}$.
